【树状数组】 HDOJ 4276 A Simple Problem with Integers

用树状数组记录所有的k的所有的余数。。。然后维护前缀和1-a 和 1-b 即可。。。

#include <iostream>  #include <queue>  #include <stack>  #include <map>  #include <set>  #include <bitset>  #include <cstdio>  #include <algorithm>  #include <cstring>  #include <climits>  #include <cstdlib>#include <cmath>#include <time.h>#define maxn 50005#define maxm 400005#define eps 1e-10#define mod 1000000007#define INF 999999999#define lowbit(x) (x&(-x))#define mp mark_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid  #define rson o<<1 | 1, mid+1, R  //typedef vector<int>::iterator IT;typedef long long LL;//typedef int LL;using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}// headint sum[maxn][10][10];int num[maxn], n, m;void init(void){for(int i = 1; i <= n; i++)for(int j = 1; j <= 10; j++)for(int k = 0; k < j; k++)sum[i][j][k] = 0;}void read(void){for(int i = 1; i <= n; i++) scanf("%d", &num[i]);scanf("%d", &m);}void updata(int x, int k, int mm, int v){for(int i = x; i <= n; i += lowbit(i)) sum[i][k][mm] += v;}int query(int x){int ans = 0;for(int i = x; i > 0; i -= lowbit(i))for(int j = 1; j <= 10; j++)ans += sum[i][j][x%j];return ans;}void work(void){int a, b, kk, k, v, x;while(m--) {scanf("%d", &kk);if(kk == 1) {scanf("%d%d%d%d", &a, &b, &k, &v);updata(a, k, a%k, v);updata(b+1, k, a%k, -v);}else {scanf("%d", &x);printf("%d\n", query(x) + num[x]);}}}int main(void){while(scanf("%d", &n)!=EOF) {init();read();work();}return 0;}